$\phi: G \to G'$ is a group homomorphism. Suppose that $|G| = 18$, $|G'| = 15$, and that $\phi$ is not the trivial homomorphism. What is the order of the kernel?
My idea is that clearly $ker\phi$ is a subgroup of G as $I \in ker\phi$; $\forall $ $x$, $y$ $\in$ $ker\phi$, $\phi(xy) = \phi(x)\phi(y) = 1$, so $xy\in ker\phi$; $\forall$ $x \in ker\phi, \phi(xx^{-1})=\phi(1) = 1 = \phi(x) \phi(x^{-1})=1\phi(x^{-1})$, so $x^{-1} \in ker(\phi).$
By the Lagrange's Theorem, |$ker \phi$| divides |$G$|, so |$ker \phi$| can just be $1$ or $2$ or $3$ or $6$ or $9$ or $18$.
As the homomorphism is not trivial, |$ker \phi$| $\neq$ 18, and as $|G'| < |G|$, |ker$\phi$| $\neq$ 1, I'm stuck then. Can I further tell among four possibilities what is $|ker\phi|$?
Best Answer
$\newcommand{\im}{\operatorname{im}}$ Notice that we have the equality $\lvert G\rvert = \lvert \ker\phi\rvert\cdot \lvert \im\phi\rvert$. Therefore, we have $\lvert \im\phi\rvert \mid {\rm gcd}(\lvert G\rvert, \lvert G'\rvert)=3$. Since $\phi$ is not trivial, it follows that $\lvert \im\phi\rvert = 3$. Consequently, $\lvert \ker\phi\rvert = 18/3=6$.